Where/Who the hell is Bryce Carlson?

[Automatic Monkey]

As QFIT has already pointed out, this is incorrect. Your 1 deck numbers are correct for a *single deck* game, but are incorrect for an 8 deck game with 1 deck remaining to be dealt and a TC 0. DUCY?

Hint: How do you know there are exactly 16 tens and 4 aces in that one deck. What if there are 20 tens and 0 aces?

And what if there are 10 tens and 10 aces? What's so special about the combination of 4 aces and 16 tens that it maximizes the number of naturals?

I hear exactly what you are saying and I know the difference between a full deck and a deck remaining with a TC of 0. But I guess "neutral deck" means TC=0 to most people. On the other hand, a counter who was counting every rank would think of it as a new and complete 52-card deck.

Now we can probably agree that the True Count Theorem applies to any kind of balanced count we care to use and that a count of, say, 4's vs. 9's would not give us any kind of information about the frequency of naturals we can expect, while a count like High-Low does. But why would the every-rank counter be able to say "I have a neutral deck with one deck remaining, therefore I am more likely to get a natural" while a High-Low or ace-neutral-count-plus-sidecount counter can't?

 

[NightStalker]

Respectfully disagree

I will repeat one more time. The floating advantage has *nothing* to do with an increased likelihood of getting dealt a blackjack. It has everything to do with basic strategy (and in some case strategy variations) being much stronger plays at deeper levels of penetration. You have more information with one deck left than you do off the top of the shoe. Specifically, in my example, you have the information that the other 7 decks were count neutral, so you know the last deck will also be count neutral. When you are playing off the top of the shoe, you do not have this luxury. You do not have any idea if that first deck out of the shoe will have a count of +10, -5 or if it will also be count neutral.

I am not certain about what you guys call 'floating advantage'. For me, floating advantage refers to effect of removal of cards in play. I agree deeper pen will produce better playing efficiency-considering cards in hand. I mostly agree with this statement.

Let's say at some point in the middle of an 8 deck shoe, someone informs you that the next 2 cards are both black. This person does not know if both are spades, both are clubs, or one of each, he only knows that both are black. Now, if the next card out is a club, what are the odds of the 2nd card also being a club? By the same logic employed by AM and NS determining the above BJ probabilities, I would assume they think the odds of the next card being a club is 0%. Since on average out of those 2 black cards, there will be one spade and one club, once they see the first card is a club, they know for certain that the second card has to be a spade. I believe this to be massively incorrect.

Wrong comparison. But I agree with your point.

In the BJ example, in a shoe game with one deck remaining and a TC of 0, when the first card is an ace, the odds of receiving a ten as a second card have actually gone down. The reverse is also true. If you receive a ten as your first card your chance of getting an ace as a second card has gone down as unlike a true single deck game, you do not know the specific composition of that remaining deck.

I am certain(I like to think that I am right, but we all are humans) that above statement is mathematically wrong. Probably you know about the game better, but there is a trivial mistake in your assumption. It's definitely a good quiz for readers? I'll restate your point in the next post, please correct me if am wrong?

 

[NightStalker]

Chap's point

BJ 6-deck shoe game with 1-deck(52cards) remaining and hi-lo TC of 0.
Now an ace is dealt. 51 cards remaining in the shoe giving Hi-lo TC = -1.

Has the probability of getting a ten decreased/increased/same after removing the ace?

 

[The Chaperone]

BJ 6-deck shoe game with 1-deck(52cards) remaining and hi-lo TC of 0.
Now an ace is dealt. 51 cards remaining in the shoe giving Hi-lo TC = -1.

Has the probability of getting a ten decreased/increased/same after removing the ace?

I'm honestly not sure if it stays the same or has gone down, but I'm pretty sure the odds of getting a ten did not go up. I'll let others think about it and weigh in.

 

[zengrifter]

I have read his book and I liked so much.
His estyle of writing is great for me.

His incorrect betting strategy skews the entire system to loser.
If he cared he would have published an errata sheet long ago. zg

 

[Alvaro]

His incorrect betting strategy skews the entire system to loser.
If he cared he would have published an errata sheet long ago. zg

I didn't know it; to what it is due to?

I dont pay any attention to "fixed betting schemes" that autors suggest (unless this were benchmarks to contrast aspect's system).

 

[zengrifter]

I dont pay any attention to "fixed betting schemes" that autors suggest (unless this were benchmarks to contrast aspect's system).

Good thinking.
Now what you should do is keep your AO2 indices but change your count tags --
-- switch the Ace and 9 tag values, changing it to ZEN.

 

[KimLee]

Yes floating advantage exists. (See http://www.blackjackincolor.com/blackjackeffects2.htm).

It just doesn't affect BJ frequencies.

That doesn't seem right. Suppose we use a 3-card deck {A,T,2}. The probability of getting a blackjack is 1/3. Now suppose we use two decks {A,A,T,T,2,2}. After dealing out three cards with a neutral count we could have {A,A,2}, or {T,T,2} in addition to {A,T,2}. Then the probability of a blackjack is not exactly the same.

I agree the probability of drawing a single ace or ten is the same, but not combination.

 

[aslan]

BJ 6-deck shoe game with 1-deck(52cards) remaining and hi-lo TC of 0.
Now an ace is dealt. 51 cards remaining in the shoe giving Hi-lo TC = -1.

Has the probability of getting a ten decreased/increased/same after removing the ace?

It has increased, assuming there is a ten among the good cards remaining.

TC -1 only tells us that the odds of getting either a ten or ace is less, not a ten alone.

But whatever the "true" probability of getting a ten was, it is now actually greater, because there are the same number of tens remaining in a fewer number of cards. If there was one ten in the remaining cards (51 to 1), now there is 1 out of 51, or 50 to 1.

 

[zengrifter]

[Probability of getting a 10] has increased, assuming there is a ten among the ... cards remaining.

I needed to re-read that one a couple times! zg

 

[aslan]

I needed to re-read that one a couple times! zg

Is that good? :laugh:

 

[zoomie]

It has increased, assuming there is a ten among the good cards remaining.

If there were at least two tens at one deck remaining and TC = 0, removing one of them increases the probability of seeing another.

 

[HockeXpert]

If there were at least two tens at one deck remaining and TC = 0, removing one of them increases the probability of seeing another.

How can removal of a card value increase the odds of seeing the same value? In its simplest form, a 2/52 chance becomes a 1/51 chance if you remove one of two tens in one deck of remaining cards.

 

[zoomie]

How can removal of a card value increase the odds of seeing the same value? In its simplest form, a 2/52 chance becomes a 1/51 chance if you remove one of two tens in one deck of remaining cards.

You are certainly correct. Maybe I was thinking of an earlier question - after pulling an ace, has the probability of drawing a ten increased? Yes, if there had been at least one remaining at TC = 0.

 

[NightStalker]

Obviously

the probability of getting a ten has increased. Is there any doubt? It was as trivial as my 4th standard geometry questions.

I was just trying to correct The Chap

 

[The Chaperone]

I agree it has increased if you knew exactly how many tens were remaining in the pack. In this case all we know is the count. I'm still not sure what the answer is, but I find your reasoning to be rather jejune.

 

[NightStalker]

The reasoning is simple:

I agree it has increased if you knew exactly how many tens were remaining in the pack. In this case all we know is the count. I'm still not sure what the answer is, but I find your reasoning to be rather jejune.

Think simple, why complicate things with the count? I am not saying that we know how many tens are remaining. I am only saying that if there is any ten in the remaining shoe, it'll increase..

Let's work out with a generic example:
Hi-lo count =0
Aaces, T tens and N neutral cards and (A+T) small cards, sounds good?
Prob of getting Ten = T/(2A+2T+N)

First card removed = Ace
Remaining composition of shoe::
(A-1)aces, T tens and N neutral cards and (A+T) small cards.. ok?
Prob of getting a Ten = T/(2A+2T+N-1)

The probability can NOT decrease in any case (Whatever composition of the deck you can come up giving hi-lo count=0). The probability will ALWAYS increase except when there are no tens remaining in the deck- then it'll remain the same =0.
In short, probability will always increase. Exception: Remains zero if no tens..

 

[k_c]

Think simple, why complicate things with the count? I am not saying that we know how many tens are remaining. I am only saying that if there is any ten in the remaining shoe, it'll increase..

Let's work out with a generic example:
Hi-lo count =0
Aaces, T tens and N neutral cards and (A+T) small cards, sounds good?
Prob of getting Ten = T/(2A+2T+N)

First card removed = Ace
Remaining composition of shoe::
(A-1)aces, T tens and N neutral cards and (A+T) small cards.. ok?
Prob of getting a Ten = T/(2A+2T+N-1)

The probability can NOT decrease in any case (Whatever composition of the deck you can come up giving hi-lo count=0). The probability will ALWAYS increase except when there are no tens remaining in the deck- then it'll remain the same =0.
In short, probability will always increase. Exception: Remains zero if no tens..

Counting system math appears to be a little different animal than the math where shoe composition consists of a known integral number of each rank.

Contrast 2 examples-

Example 1
26 cards remain from a single deck consisting of 2 each (A,2,3,4,5,6,7,8,9) and 8 tens.

Prob(A,2,3,4,5,6,7,8,9) = 2/26 = 1/13 = .0769231
Prob(T) = 8/26 = 4/13 = .307962

An ace is removed leaving 25 cards and probs below
Prob(A) = 1/25 = .04
Prob(2,3,4,5,6,7,8,9) = 2/25 = .08
Prob(T) = 8/25 = .32

Specifically removing an ace results in these changes:
Prob(A) decreased from .0769231 to .04
Prob(2,3,4,5,6,7,8,9) increased from .0769231 to .08
Prob(T) increased from .307962 to .32

An ace was specifically removed and that was the only rank that decreased in probability.
Ranks other than ace increased in probability.

The probs in example 1 are what are used in a combinatorial analyzer where calculations are made based upon exactly how many of each rank is present in a shoe.

Example 2
26 cards remain from a single deck and HiLo running count is 0. It turns out that regardless of number of starting decks that for a HiLo RC=0 and 1/2 shoe remaining
Prob(A,2,3,4,5,6,7,8,9) = 1/13
Prob(T) = 4/13
provided all that is known is RC=0 (nothing specifically removed)
For number of remaining cards other than half shoe (or full shoe) the probs are generally close to 1/13 and 4/13 but aren't exactly 1/13 and 4/13.

Below shows starting probs for a single deck, 26 cards remaining, RC=0, nothing specifically removed and probs for a single deck, 25 cards remaining, RC=-1, 1 ace specifically removed.

single deck, 26 cards remaining, RC=0, nothing specifically removed
Count tags {1,-1,-1,-1,-1,-1,0,0,0,1}
Decks: 1
Cards remaining: 26
Initial running count (full shoe): 0
Running count: 0
Specific removals
A: 0
2: 0
3: 0
4: 0
5: 0
6: 0
7: 0
8: 0
9: 0
T: 0

Number of subsets for 26 cards: 231
Prob of running count 0 from 1 deck: 0.124165

p[1] 0.0769231 p[2] 0.0769231 p[3] 0.0769231 p[4] 0.0769231 p[5] 0.0769231
p[6] 0.0769231 p[7] 0.0769231 p[8] 0.0769231 p[9] 0.0769231 p[10] 0.307692

single deck, 25 cards remaining, RC=-1, 1 ace specifically removed
Count tags {1,-1,-1,-1,-1,-1,0,0,0,1}
Decks: 1
Cards remaining: 25
Initial running count (full shoe): 0
Running count: -1
Specific removals
A: 1
2: 0
3: 0
4: 0
5: 0
6: 0
7: 0
8: 0
9: 0
T: 0

Number of subsets for 25 cards: 230
Prob of running count -1 from 1 deck: 0.124165

p[1] 0.0572156 p[2] 0.0804731 p[3] 0.0804731 p[4] 0.0804731 p[5] 0.0804731
p[6] 0.0804731 p[7] 0.078423 p[8] 0.078423 p[9] 0.078423 p[10] 0.30515

Specifically removing an ace in example 2 results in these changes:
Prob(A) decreased from .0769231 to .0572156
Prob(2-6) increased from .0769231 to .0804731
Prob(7-9) increased from .0769731 to .078423
Prob(T) decreased from .307692 to .30515

In example 2 when cards are specifically removed from a shoe where the probabilities of drawing each rank are computed from RC/remaining cards the dynamics are different than in example 1. A rank specifically removed not only reduces the prob of drawing that rank on the next draw but also reduces the prob of drawing a rank in the counting system card group to which it belongs.

Bottom line is that when you start by listing the exact composition that results in a given count you are reverting to example 1 and ignoring all of the other possible subsets that could have comprised the same count which are considered in example 2.

 

[NightStalker]

Wrong assumption results in wrong conclusion

You are totally mistaken in your comparison. This is my last try, I hope you can see the trivial things. What about people using only Ace-Five count where 10's=0?

your mistake is::

Example 2
Prob(A,2,3,4,5,6,7,8,9) = 1/13
Prob(T) = 4/13

p[1] 0.0769231 p[2] 0.0769231 p[3] 0.0769231 p[4] 0.0769231 p[5] 0.0769231
p[6] 0.0769231 p[7] 0.0769231 p[8] 0.0769231 p[9] 0.0769231 p[10] 0.307692

You are trying to tell: that HL RC of -1 has P(T) less than P(T) ar RC 0, everyone knows it. You are comparing two averages. But we are talk about average of the difference. This is what you had done in example two::

Before Ace, All subset:: Average P(T)
After Ace, All subset:: Average P(T)'
Delta= Average[P(T)'] - Average[P(T)]
Now you are saying that delta is -ive, which is a wrong approach.

Ideal approach or point of discussion is::

In each subset, find delta and then give the average of Delta.

We were discussing on
Average[P(T')-P(T)] > = 0
and it is and it'll be always in EVERY POSSIBLE SUBSET
where P(T)= probability of Ten in random shoe of hi-lo RC=0
And P(T') = probability of Ten after an ACE have been removed from above random shoe.
P(T') != probability of Ten after an ACE have been removed from ANY random shoe with hi-lo RC=0.

 

[k_c]

You are totally mistaken in your comparison. This is my last try, I hope you can see the trivial things. What about people using only Ace-Five count where 10's=0?

your mistake is::


You are trying to tell: that HL RC of -1 has P(T) less than P(T) ar RC 0, everyone knows it. You are comparing two averages. But we are talk about average of the difference. This is what you had done in example two::

Before Ace, All subset:: Average P(T)
After Ace, All subset:: Average P(T)'
Delta= Average[P(T)'] - Average[P(T)]
Now you are saying that delta is -ive, which is a wrong approach.

Ideal approach or point of discussion is::

In each subset, find delta and then give the average of Delta.

We were discussing on
Average[P(T')-P(T)] > = 0
and it is and it'll be always in EVERY POSSIBLE SUBSET
where P(T)= probability of Ten in random shoe of hi-lo RC=0
And P(T') = probability of Ten after an ACE have been removed from above random shoe.
P(T') != probability of Ten after an ACE have been removed from ANY random shoe with hi-lo RC=0.

You are correct for A-5 count. Remove an ace and prob of all other ranks increases. Please note that prob of 5 has increased more than the other ranks, This is the nature of counts. Some cards are given more weight than others.

Count tags {1,0,0,0,-1,0,0,0,0,0}
Decks: 1
Cards remaining: 26
Initial running count (full shoe): 0
Running count: 0
Specific removals
A: 0
2: 0
3: 0
4: 0
5: 0
6: 0
7: 0
8: 0
9: 0
T: 0

Number of subsets for 26 cards: 25
Prob of running count 0 from 1 deck: 0.270528

p[1] 0.0769231 p[2] 0.0769231 p[3] 0.0769231 p[4] 0.0769231 p[5] 0.0769231
p[6] 0.0769231 p[7] 0.0769231 p[8] 0.0769231 p[9] 0.0769231 p[10] 0.307692


Count tags {1,0,0,0,-1,0,0,0,0,0}
Decks: 1
Cards remaining: 25
Initial running count (full shoe): 0
Running count: -1
Specific removals
A: 1
2: 0
3: 0
4: 0
5: 0
6: 0
7: 0
8: 0
9: 0
T: 0

Number of subsets for 25 cards: 25
Prob of running count -1 from 1 deck: 0.270528

p[1] 0.0496287 p[2] 0.0782493 p[3] 0.0782493 p[4] 0.0782493 p[5] 0.0896287
p[6] 0.0782493 p[7] 0.0782493 p[8] 0.0782493 p[9] 0.0782493 p[10] 0.312997